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3.1 — The In-Betweeners
Core idea When an exponential quantity doubles in one unit of time, it must grow by the same proportional factor over every equally long subdivision. Equal time intervals do not add equal amounts.
1. In class: A Develop Understanding Task
Begin with growth at whole-number inputs
Suppose a quantity starts at 4 and doubles whenever the input increases by 1. The function is \(f(x)=4\cdot 2^x\). The 4 is the initial value; the 2 is the growth factor for one full unit of input.
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(f(x)\) | 4 | 8 | 16 | 32 | 64 |
Given \(f(x)=4\cdot 2^x\), calculate \(f(3)\):
\[f(3)=4\cdot2^3=4\cdot8=32.\]
Move to the next whole-number input by multiplying 32 by 2, giving \(f(4)=64\). Moving backward from 4 to 3 reverses the multiplication: divide by 2.
A continuous exponential function is defined at inputs between whole numbers as well. The question is how to assign values between 4 and 8, between 8 and 16, and so on, while preserving the same pattern of proportional growth.
Why ordinary midpoint averaging is not exponential growth
Travis proposes averaging two consecutive outputs. Between \(f(0)=4\) and \(f(1)=8\), he obtains \((4+8)/2=6\). His values halfway between 8 and 16, and between 16 and 32, would be 12 and 24.
| Input | Start | Halfway by averaging | End |
|---|---|---|---|
| 0 to 1 | 4 | 6 | 8 |
| 1 to 2 | 8 | 12 | 16 |
| 2 to 3 | 16 | 24 | 32 |
| 3 to 4 | 32 | 48 | 64 |
The two half-interval factors between 4 and 8 would be \(6/4=1.5\) and \(8/6=4/3\). They are different. Thus Travis’s rule does not produce a consistent exponential growth factor over equal half-unit intervals.
A linear function adds the same amount whenever x rises by 1. For \(g(x)=4+4x\), we have \(g(0)=4\), \(g(1)=8\), and \(g(1/2)=6\). The average is correct for this linear function because each half-step adds 2.
For a doubling exponential function, an equal half-step must multiply by the same factor instead.
Find the multiplier for a half-step
Let \(r\) be the factor for every increase of \(1/2\) in x. Two half-steps make one whole step. Therefore two multiplications by \(r\) must equal multiplication by 2:
We choose the positive number because the model describes positive growth. The square root of 2 is the positive number whose square is 2.
The halfway output is the geometric mean of the two endpoint outputs: for example, \(4\sqrt2\approx5.657\), which is less than Travis’s 6.
Start from 8 at \(x=1\). One half-step later, multiply by \(\sqrt2\): \(8\sqrt2\approx11.314\). Another half-step later, multiply by \(\sqrt2\) again:
\[8\sqrt2\cdot\sqrt2=8\cdot2=16.\]
The same factor works for both intervals.
| \(x\) | 0 | \(1/2\) | 1 | \(3/2\) | 2 | \(5/2\) | 3 | \(7/2\) | 4 |
|---|---|---|---|---|---|---|---|---|---|
| Exact \(f(x)\) | 4 | \(4\sqrt2\) | 8 | \(8\sqrt2\) | 16 | \(16\sqrt2\) | 32 | \(32\sqrt2\) | 64 |
| Approx. | 4 | 5.657 | 8 | 11.314 | 16 | 22.627 | 32 | 45.255 | 64 |
Writing \(2^{1/2}\) is another way of labeling the half-step multiplier \(\sqrt2\) in this context: \(2^{1/2}\cdot2^{1/2}=2\). A later lesson develops the general rules for fractional exponents; here we need only understand the consistent half-step factor.
Find the multiplier for a third-step
To split a one-unit doubling into three equal steps, choose a positive factor \(s\) that works three times in a row:
The cube root of 2 is the positive number whose cube is 2. Three consecutive third-steps give \(\sqrt[3]{2}^3=2\).
Starting from \(f(0)=4\), the outputs at \(x=1/3,2/3,1\) are:
\[4\sqrt[3]{2}\approx5.040,\qquad4(\sqrt[3]{2})^2\approx6.350,\qquad4(\sqrt[3]{2})^3=8.\]
Each step multiplies by exactly the same factor, approximately 1.2599.
| \(x\) | 0 | \(1/3\) | \(2/3\) | 1 | \(4/3\) | \(5/3\) | 2 | \(7/3\) | \(8/3\) | 3 |
|---|---|---|---|---|---|---|---|---|---|---|
| \(f(x)\) approx. | 4.000 | 5.040 | 6.350 | 8.000 | 10.079 | 12.699 | 16.000 | 20.159 | 25.398 | 32.000 |
Estimating a value such as \(f(5/3)\)
Since \(5/3=1+2/3\), start at \(f(1)=8\) and take two third-steps. Multiply by \(\sqrt[3]{2}\) twice: \[f(5/3)=8(\sqrt[3]{2})^2\approx8(1.5874)\approx12.699.\] This also matches extending the third-step table.
A useful pattern to remember for positive exponential growth: when a whole unit multiplies by \(b\), two equal substeps need a factor whose square is \(b\); three equal substeps need a factor whose cube is \(b\).
See why the exponential curve bends
The graph becomes steeper because multiplying a larger amount by the same factor produces a larger numerical increase. For example, doubling 4 adds 4, while doubling 32 adds 32. A straight line would have a constant additive rate instead.
2. Ready: arithmetic and geometric sequences (problems 1–11)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A sequence lists values in order. To classify it, compare consecutive terms. A constant difference makes an arithmetic (additive) sequence; a constant ratio makes a geometric (multiplicative) sequence.
| Feature | Arithmetic sequence | Geometric sequence |
|---|---|---|
| Same operation | Add a fixed number \(d\) | Multiply by a fixed factor \(r\) |
| First term \(a\); term number \(n\) | \(a_n=a+(n-1)d\) | \(a_n=a\cdot r^{n-1}\) |
| Example | 5, 9, 13, 17, … | 5, 10, 20, 40, … |
The \(n-1\) is important: the first term requires zero changes, the second term one change, and the third term two changes.
The sequence 13, 8, 3, −2, … subtracts 5 each time, so \(a=13\), \(d=-5\).
\[a_n=13+(n-1)(-5)=18-5n.\]
Check: \(a_1=13\), \(a_4=-2\).
The sequence 2, −6, 18, −54, … multiplies by −3 each time. Therefore \(a_n=2(-3)^{n-1}\). The signs alternate because the multiplier is negative. This is a discrete sequence rule; it does not describe the positive continuous growth curves of today’s main task.
Read an exponential graph accurately
For Ready problems 6–11, read \(f(2)\) by moving vertically from input 2 until reaching the curve, then horizontally to the output axis. For \(f(x)=4\), start at height 4 and move horizontally to the curve before reading x. These are different questions: given x, find y versus given y, find x.
Suppose the graph contains \((0,1),(2,2),(4,4),(6,8),(8,16)\). Doubling takes two units of x, so each one-unit increase uses the factor \(\sqrt2\). The graph’s rule is \(f(x)=2^{x/2}\), equivalently \(f(x)=(\sqrt2)^x\).
Thus \(f(2)=2\), \(f(6)=8\), and \(f(x)=4\) at \(x=4\). The graph in the packet is intended to represent this approximate pattern; check its axis marks carefully.
Check your work — Ready 1–11 answer key
- a. Geometric, ratio 2. Missing values: 64, 128, 256. b. \(f(n)=2^n\), \(n\ge1\).
- a. Arithmetic, difference −16. Missing values: 2, −14, −30, −46. b. \(f(n)=66-16(n-1)\).
- a. Geometric, ratio −3. Missing values: −243, 729, −2187, 6561. b. \(f(n)=(-3)^n\).
- a. Geometric, ratio \(\frac12\). Missing values: \(10,5,\frac52,\frac54\). b. \(f(n)=160(\frac12)^{n-1}\).
- a. Arithmetic, difference 7. Missing values: 19, 26, 33, 40. b. \(f(n)=-9+7(n-1)\).
- \(f(2)=2\).
- \(x=4\).
- \(f(6)=8\).
- \(x=8\).
- The output doubles whenever the input increases by 2; each one-unit increase multiplies the output by \(\sqrt2\).
- \(f(x)=2^{\frac{x}{2}}\).
3. Set: growth tables with halves and thirds (problems 12–14)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The bacteria tables give the starting quantity and the multiplier for one whole year. The blanks include fractional years. Find the positive equal-interval factor first, and use it repeatedly across the row.
Start with 2 bacteria units at year 0, and triple in a full year. A half-year factor \(r\) must satisfy \(r^2=3\), so \(r=\sqrt3\).
\[B(1/2)=2\sqrt3\approx3.464,\quad B(1)=2(\sqrt3)^2=6,\quad B(3/2)=6\sqrt3\approx10.392.\]
The year-2 value is \(6\cdot3=18\). Continue multiplying by \(\sqrt3\) for every additional half-year.
If a full year multiplies by 3, the factor for each third of a year is \(s=\sqrt[3]{3}\approx1.44225\).
At \(t=2/3\), multiply twice from the initial value: \[B(2/3)=2s^2=2(\sqrt[3]{3})^2\approx4.160.\]
After three third-steps, \(B(1)=2s^3=6\).
If the full-year multiplier is 4, then the half-year factor is the positive number whose square is 4: \(r=2\).
From an initial amount of 2, the first four half-year values are \(2,4,8,16,32\) at years \(0,1/2,1,3/2,2\). In this case every half-year entry is a whole number.
The title says “Evaluate the Expressions with Rational Exponents,” but the exercises actually present growth tables. The intended work is to determine a constant half-year or third-year multiplier and use it to fill the table. Exact radical forms and reasonable decimal approximations express the same model. The tables describe modeled quantities, so fractional numerical predictions need not be whole-number bacterial counts.
Check your work — Set 12–14 answer key
- Missing outputs, in column order: \(2\sqrt3,6\sqrt3,18,18\sqrt3,54,54\sqrt3,162\).
- Missing outputs, in column order: \(2\sqrt[3]3,2\sqrt[3]9,6\sqrt[3]3,6\sqrt[3]9,18,18\sqrt[3]3,18\sqrt[3]9\).
- Missing outputs, in column order: 4, 16, 32, 64, 128, 256, 512.
4. Go: simplify expressions with integer exponents (problems 15–20)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
These homework items review laws for integer exponents. They do not require the general fractional-exponent rules that the next section develops. Exponent rules assume the stated base is nonzero wherever division is involved.
| Situation | Rule | Reason / example |
|---|---|---|
| Multiply same base | \(a^m a^n=a^{m+n}\) | \(3^2\cdot3^4=3^6\) |
| Divide same base | \(a^m/a^n=a^{m-n}\) | \(5^7/5^2=5^5\) |
| Exponent zero | \(a^0=1\) for \(a\ne0\) | \(11^0=1\) |
| Negative integer exponent | \(a^{-n}=1/a^n\) | \(2^{-4}=1/2^4\) |
Problem: Simplify \(6^3\cdot6^4\) in exponential form.
Step 1. The bases are both 6. Add the exponents when multiplying equal bases.
\[6^3\cdot6^4=6^{3+4}=\boxed{6^7}.\]
Check: Three factors of 6 and four more factors of 6 give seven factors total. Do not multiply the exponents.
Problem: Simplify \(\dfrac{5^{-2}\cdot5^6}{5^3}\) in exponential form.
Step 1. Combine powers across the numerator: \(5^{-2}\cdot5^6=5^4\).
Step 2. Divide powers by subtracting the denominator exponent: \[\frac{5^4}{5^3}=5^{4-3}=\boxed{5^1}.\]
Common error: A negative exponent does not make the value negative; \(5^{-2}=1/25\).
Problem: Simplify \(\dfrac{7^0\cdot7^{-3}}{7^{-5}}\) in exponential form.
\[\frac{7^{0+(-3)}}{7^{-5}}=7^{-3-(-5)}=\boxed{7^2}.\]
Subtracting a negative exponent means adding. Notice also that \(7^0=1\).
Check your work — Go 15–20 answer key
- \(3^7\).
- \(5^1\).
- \(\frac1{2^5}\).
- \(17^0=1\).
- \(7^2\).
- \(3^{-4}\).
5. Additional learning resources
These are optional supports for this specific daily lesson. All links require internet access.
| Resource and direct link | Best use for Section 3.1 | Scope / What to skip |
|---|---|---|
| Paul’s Online Math Notes — Basic Exponential Functions | Written discussion of exponential values, tables and graph behavior; supports the main task and Ready graph reading. | Some extra material. Concentrate on basic exponential examples. Skip material about \(e\) and calculus. |
| Paul’s Online Math Notes — Exponential Functions Practice | Extra function-evaluation and graph practice with available solutions. | Some extra material. Use evaluation questions, particularly fractional inputs; skip transformed curves and natural-base examples. |
| OpenStax — 1.3 Radicals and Rational Exponents | Optional reference on square/cube roots if half- and third-step multipliers are unfamiliar. | Significantly broader. Use only the opening explanations of square and cube roots; leave general fractional exponent manipulations and radical simplification for later sections. |
Teacher / parent review: serious packet issues
Section 3.1, Set, problems 12–14: The heading “Evaluate the Expressions with Rational Exponents” does not match the actual table-completion questions. The instruction describes growth but does not explain whether the resulting bacteria entries should be reported exactly or approximately. Suggested revision: “For each table, find the constant growth factor for each indicated fractional year. Fill the entries using exact expressions or approximations rounded to three decimal places.”
Section 3.1, Ready, problems 6–11: Reading exact values from the printed graph requires interpreting a lightly labeled graph; students may give slightly different estimates. Suggested revision: Clearly label the unit of each horizontal grid step and indicate that values read from the graph are approximate unless exact coordinate points are supplied.
No further serious mathematical errors were identified in the reviewed section.
Coverage of the original Section 3.1 packet
| Packet component | Requested skill | Where taught |
|---|---|---|
| Main task 1–2 | Whole-number doubling table and graph | In class, In class |
| Main task 3–4 | Evaluate Travis’s midpoint approach; contrast linear and exponential patterns | In class, In class |
| Main task 5, 7 | Geometric half-step multiplier and interpolated values | In class |
| Main task 6, 8–9 | Third-step multiplier and estimating \(f(5/3)\) | In class |
| Ready 1–5 | Classify arithmetic/geometric sequences and write explicit rules | Section 2 |
| Ready 6–11 | Read function inputs/outputs from graph and find explicit exponential rule | Section 2 |
| Set 12–14 | Tripling/quadrupling growth tables at fractional years | Section 3 |
| Go 15–20 | Product, quotient, zero, and negative integer exponent rules | Section 4 + original worked Go problems |
Source curriculum: Mathematics Vision Project, Secondary Math II, Module 3, Section 3.1: The In-Betweeners and its Ready, Set, Go homework (2017), distributed under CC BY 4.0. This supplement is independently authored.
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